MathOverflow Asked by User3568 on October 7, 2020

Let $X$ be a non-singular complex variety with a big line and base point free bundle $M$ on it. My question is can we say that for any locally free sheaf $F$ on $X$, $F otimes M^n$ is globally generated for $n gg 0$.

Motivation: If $M$ were an ample line bundle then all we need is that $F$ is coherent sheaf. But since we are given a much stronger condition on $F$ (which is local freeness) can we say the same thing with $M$ just being big and base point free.

I tried to use the fact that any big line bundle is tensor product of an ample line bundle and an effective line bundle.

I am not even sure that this has to be true but am unable to find a counterexample.

In general the answer is no, even $F$ is a line bundle itself. It is easy to see that a globally generated line bundle is nef, and if $F$ is not nef, and the segment between $F$ and $M$ does not intersect with the ample cone in $N^{1}(X)$, then $F otimes M^{n}$ is numerically propotional to a divisor lies in the interior of the segment, thus is not nef.

Correct answer by Zhengyu Hu on October 7, 2020

In my view being big corresponds to being "generically ample", that is being ample outside a closed subset. This closed subset is called the augmented base locus of $M$ and it is often denoted by $mathbb{B}_+(M)$ ( see http://de.arxiv.org/abs/math/0308116v2.pdf). So, if for example you take $M$ big and $F$ to be a line bundle you can prove that $M^motimes F$ is always globally generated (i.e. base point free) outside the augmented base locus of $M$ (note that you don't need global generation of $M$ for this). In the example of Anton Geraschenko in fact $mathbb{B}_+(M)$ is exactly the exceptional divisor. Something similar might hold if you consider locally free sheaves of higher rank.

Answered by Gianni Bello on October 7, 2020

Here is a simple counterexample (of the form Zhengyu Hu suggested). Take $X$ to be the blowup of a point in $mathbb P^2$, $M$ the pullback of $mathcal O(1)$ under the blowup map, and $F$ the line bundle associated to the exceptional divisor $E$. Sections of $Fotimes M^n$ are rational functions which may have a pole of order 1 along $E$ and a pole of order up to $n$ along a line not meeting $E$. However, any rational function $f$ actually having a pole along $E$ (i.e. generating $Fotimes M^n$ at the points of $E$) must have a pole along some other divisor passing through $E$ (since $X$ has the same rational functions as $mathbb P^2$).

Answered by Anton Geraschenko on October 7, 2020

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